Question

Hello please help me with these questions

Answer 1

Explanation:

20. the Circle with the center is at (4, -3) and the radius is r=8

Circle equation.

`\left(x−a\right)^2+\left(y−b\right)^2=r^2\:\:\mathrm{is\:the\:circle\:equation\:with\:a\:radius\:r,\:centered\:at} (a,b)`

`\mathrm{Rewrite}\:\left(x-4\right)^2+\left(y+3\right)^2=64\:\mathrm{in\:the\:form\:of\:the\:standard\:circle\:equation}`

`\left(x-4\right)^2+\left(y-\left(-3\right)\right)^2=8^2`

And so the circle properties are:
`\left(a,\:b\right)=\left(4,\:-3\right),\:r=8`

21.
`\mathrm{TheCircle ..at..the \:\:center..is \:at}\:\left(-2,\:0\right)\:\mathrm{and\:radius}\:r=√(13)`

Circle equation:

`\left(x−a\right)^2+\left(y−b\right)^2=r^2\:\:\mathrm{is\:the\:circle\:equation\:with\:a\:radius\:r,\:centered\:at}\:\left(a,\:b\right)`

`\mathrm{Rewrite}\:\left(x+2\right)^2+y^2=13\:\mathrm{in\:the\:form\:of\:the\:standard\:circle\:equation}`

`\left(x-\left(-2\right)\right)^2+\left(y-0\right)^2=\left(√(13)\right)^2`

`\left(a,\:b\right)=\left(-2,\:0\right),\:r=√(13)`